As a follow up to Wednesday’s sad balloon post, the repair that lofted it back to the ceiling was a temporary reprieve, unsurprisingly. After 24 hours, more or less, it had sunk back down to the point where the ribbon was just barely touching the floor.

On the one hand, it looks kind of pathetic again. On the other hand, though, this is a chance to do some more physics. See, I know how much ribbon I clipped off the end to get it to lift off again, and if it’s fallen back to where it’s just touching the floor, that means that the buoyant force due to the displaced air has decreased by an amount equal to the weight of that much ribbon.

So, if we work out how much helium is required to lift that weight, we know how much helium leaked out overnight. Which tells us how quickly the helium inside leaked through the Mylar of the balloon, which is kind of a cool thing to be able to measure with such a simple experiment.

So, how much helium do we need? Well, the net buoyant force is equal to the weight of the displaced air minus the weight of the helium that replaces it. Air has a density of 1.29 kg/m^{3}, while helium has a density of 0.18 kg/m^{3}, so a balloon full of a cubic meter of helium at atmospheric pressure would produce a lift force of:

F=(weight of air)- (weight of helium) = (Volume)((air density)-(He density))(

g)

F= (1m^{3})((1.29 kg/m^{3})-(0.18 kg/m^{3}))(9.8 N/kg) = 11N

The volume of the balloon is nowhere near a cubic meter, of course, but we can just call this 11 N/m^{3}, and use this to find the volume of helium from the change in lift.

So, how much did the lift change? I could take the ribbon to campus and weigh it on the milligram scale, but that would feel like cheating, so let’s estimate it from the dimensions. I cut 11 in (about 28 cm) of ribbon off to get the balloon to float up to the ceiling again, and the ribbon is about 0.5cm wide. The thickness is too small to measure with the rulers I have here, but I folded it over on itself until there were ten bits of ribbon stacked up, and that stack was about 0.1cm high, so we’ll say that it’s 00.01cm thick (that’s 100 microns, or about the thickness of a human hair, which is probably a little small, but close enough).

If you’re playing along at home, that works out to a total volume of 0.14 cm^{3}, or 0.00000014 m^{3}. To get a mass from this, we need a density. I’m not sure what the ribbon actually it, but it’s sort of similar to paper, so let’s use the density of paper, which the always authoritative Wikipedia gives as 800 kg/m^{3}. Put all that together, and you have an estimated mass of 0.000112 kg, or about a tenth of a gram.

The weight of that length of ribbon, then, is that mass multiplied by the strength of gravity, which is 0.0011 N. Dividing that by our figure of 11 N/m^{3} for the lifting power of helium suggests that the balloon lost about 9.9×10^{-5} m^{3}, or about 100 ml.

That seems a little big, but then, it’s not all that much compared to the starting volume of the balloon. When spread out, the balloon has a circular cross-section with a radius of about 20cm, and inflated to about 10cm thick, giving a total volume of 13,000 cm^{3}. If it loses 100 cm^{3} of helium a day, it would take 130 days for *all* of the helium to leave (assuming, of course, that the rate is constant.

Now, I probably overestimated the volume of the balloon by considering it as a cylinder. But then, I probably overestimated the density of the ribbon, too, and those two effects will tend to cancel each other out. So this is probably in the right ballpark: a hundred-day lifetime for the complete loss of helium from a Mylar balloon.

That might seem awfully long, but consider this: That particular balloon is left over from *my* birthday, back in June. So it’s already survived 60 days with at least some of its original helium.

(Of course, this requires the balloon to be treated with respect, or at least benevolent neglect. A similar balloon from SteelyKid’s birthday has already lost much more of its helium, but that was a consequence of an enthusiastic toddler hugging and bouncing it around for a few days.)

So, there’s your latest dose of physics-related overthinking of simple objects…